Let's imagine there's a block of mass m sitting on a rough surface with a kinetic friction coefficient $\mu$. It's being pulled with a constant force $A$ at $h$ degrees above the horizontal and is displaced a distance $d$. The work done by $A$ on the block is positive and is the horizontal component of $A$ times the displacement:

$$A\cos (h)*d.$$

I thought this meant that the work done by friction would be either the negative of that amount or else

$$\mu*(mg-A\sin (h))*d$$

but apparently not?

  • $\begingroup$ What does it mean if the force and the displacement are in opposite directions? $\endgroup$
    – BowlOfRed
    Commented Oct 23, 2015 at 5:07
  • $\begingroup$ It means the work done is negative. $\endgroup$
    – AaronF
    Commented Oct 23, 2015 at 14:06

1 Answer 1


You were very close. The work done by the constant force of kinetic friction is W_fric = Fdcos(a) where a is the angle between the friction and the displacement. Kinetic friction always points in the direction opposite the motion, so a equals 180°. This was your error. If the force and displacement point in opposite directions the angle bewteen them is 180° not 0°.

F = un = u(mg-Asin(h)) (as you have). d=d (highly insightful). And cos(a) = cos(180°) = -1. So W_fric = -u*(mg-Asin(h))*d, which is really just the negative of your answer. Also as a general rule, kinetic friction always points opposite to the direction of motion and hence always does negative work. Thanks for the question, I hoped my answer helped you out, and have a nice day.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.